Problem: We know that $-\frac{2}{1-2x}=-2-4x-8{{x}^{2}}-16{{x}^{3}}-...-2{(2x)}^{n-1}+...$ for $x\in\left(-\dfrac{1}{2},\dfrac{1}{2}\right)$. Using this fact, find the function that corresponds to the following series. $ -2x-2x^2-\frac{{8}}{3}x^3-4x^4-...$ Choose 1 answer: Choose 1 answer: (Choice A) A $ \arctan \left( 1+2x \right)$ (Choice B) B $\arctan \left( 1-2x \right)$ (Choice C) C $-\arctan \left( 1-2x \right)$ (Choice D) D $\ln \left( 1+2x \right)$ (Choice E) E $-\ln \left( 1-2x \right)$ (Choice F) F $\ln \left( 1-2x \right)$
Explanation: First, note that the derivative of $-2x-2x^2-\frac{{8}}{3}x^3-4x^4-...$ is $-2-4x-8{{x}^{2}}-16{{x}^{3}}-...-2{(2x)}^{n-1}+...=-\frac{2}{1-2x}$ Then $\int{\left(-\frac{2}{1-2x}\right)dx}=\int{\left(-2-4x-8x^2-16x^3-...-2(2x)^{n-1}\right)dx}$ Using the $~u$ -substitution $~u=1-2x~$ with $~du=-2dx\,$, we have $\int{-\frac{2}{1-2x}}\text{ }dx=\ln \left( 1-2x \right)+C$ Therefore $\ln(1-2x)+C=-2x-2x^2-\frac{8}{3}x^3-4x^4-...$ for some value of $~C\,$. Now let $~x=0~$ and equate the right-hand sides of the last equation to see that $~C=0\,$. Thus, $ -2x-2x^2-\frac{{8}}{3}x^3-4x^4-...=\ln \left( 1-2x \right)\,$